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3 years ago

KN is perpendicular bisector of MQ identify the value of x

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

4 0

Answer:

x = 6

Step-by-step explanation:

Since KN is the perpendicular bisector, that means ∠KNM = ∠KNQ = 90° and MN = NQ so therefore, since they are right triangles, ΔKNM ≅ ΔKNQ because of HL. Therefore, KM = KQ by CPCTC so:

5x - 3 = 3x + 9

2x = 12

x = 6

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Please help. Just put the coords as the answer.

Julli [10]

Answer:

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3 years ago

The dolphins scored 2x - 7 points, while the jaguars scored -5x - 3 points. how many more points did the dolphins score than the

mr Goodwill [35]

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3 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra

Nana76 [90]

The length of BC is $4 \sqrt{5}$ .

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

<u>Leg rule of geometric mean theorem:</u>

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

$\Rightarrow 16\times 5 = x\times x$

$\Rightarrow 80= x^2$

$\Rightarrow 16\times 5= x^2$

Taking square root on both sides.

$\Rightarrow \sqrt{16\times 5} = \sqrt{x^2}$

$\Rightarrow \sqrt{4^2\times 5} = \sqrt{x^2}$

square and square roots get canceled, we get

$\Rightarrow 4\sqrt{ 5} = x$

The length of BC is $4 \sqrt{5}$ .

7 0

3 years ago

Please help ASAP 40 pts

natali 33 [55]

Answer:

x=2

Step-by-step explanation:

We know AC = BC since this is an isosceles triangle

6 = 2x+2

Subtract 2 from each side

6-2 =2x+2-2

4 =2x

Divide each side by 2

4/2 =2x/2

2=x

4 0

3 years ago

Read 2 more answers

The function of (x) varies directly with x^2, and f(x)=96 when x=4 what is the value of f(2)

makkiz [27]

$\bf \qquad \qquad \textit{direct proportional variation}
\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------$

$\bf \textit{f(x) varies directly with }x^2\qquad f(x)=kx^2
\\\\\\
\textit{we also know that }
\begin{cases}
f(x)=96\\
x=4
\end{cases}\implies 96=k(4)^2\implies 96=16k
\\\\\\
\cfrac{96}{16}=k\implies 6=k\qquad therefore\qquad \boxed{f(x)=6x^2}
\\\\\\
\textit{now, when }\stackrel{f(2)}{x=2}\textit{ what is \underline{f(x)}?}\qquad f(2)=6(2)^2$

3 0

3 years ago